Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Scanner;

/**
 * Coins
 */
public class Main {

    public static void main(String[] args) throws Exception {
        try (Scanner sc = new Scanner(System.in)) {
            int N = sc.nextInt();
            double[] P = new double[N + 1];
            for (int i = 1; i <= N; i++) {
                P[i] = sc.nextDouble();
            }

            double[][] dp = new double[N + 1][N + 1];
            dp[0][0] = 1;

            for (int i = 1; i <= N; i++) {
                for (int j = 0; j <= i; j++) {
                    dp[i][j] = dp[i - 1][j] * (1 - P[i]);
                    if (j > 0) {
                        dp[i][j] += dp[i - 1][j - 1] * P[i];
                    }
                }
            }

            double sum = 0;
            for (int i = N / 2 + 1; i <= N; i++) {
                sum += dp[N][i];
            }

            System.out.println(sum);
        }
    }

}